www.rtmj.net > 已知等差数列{An}公差不为0,且A3=5,A1,A2,A5成等比数列.(1)求数列{An}的通项公...

已知等差数列{An}公差不为0,且A3=5,A1,A2,A5成等比数列.(1)求数列{An}的通项公...

已知等差数列{an}公差不为0,且a3=5,a1,a2,a5成等比数列.(1)求数 答:(1

数学知识都还给老师了哈哈,回答你第一问。 a1+a2=2a1+d=1 a2,a3,a5成等比数劫

(1)由S3=9,可得3a1+3d=9即a1+d=3①(2分)∵a1,a2,a5成等比数列.∴a1(

(Ⅰ)设等差数列的公差是d,∵a1,a2,a5成等比数列,即2,2+d,2+4d成等比数列,∴(2+

解答:(1)解:设公差为d,则d≠0,又a1,a2,a4成等比数列,则有a22=a1a4,又首项a1

1、(a1+d)/a1=(a1+4d)/(a1+d) 解得d=0或1 由题得d=1 an=n-

(Ⅰ)由题意知S3=9a22=a1a5即3(a1+d)=9(a1+d)2=a1(a1+4d).所以a

∵等差数列{an}的公差和首项都不等于0,且a2,a4,a8成等比数列,∴a42=a2a8,∴(a1

(本小题满分12分)解:(Ⅰ)设数列{an}的公差为d(d≠0),∵S3=a4+2,∴3a1+3×2

(Ⅰ)设{an}的公差为d,(d≠0),∵a1,a2,a5成等比数列,∴a22=a1?a5(2分)又

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